Combo Pack -II

1) Total no. of stereoisomers for-

Me--CH(Cl)--CH=CH--CH=C=C(Me)--Me .....?

2) Total no. of products on MOnochlorination of n-Pentane in the presence of sunlight ?

3)Arrange in Decreasing order of Rate of ElectroPhillic Substitution in the presence of HBr-

I- (Ph)2--C=CH2
II- Ph--CH-CH2
III- Ph--CH=CH--Ph
IV- MeO--CH=CH2

4) For the Complex- [Fe(H2O)5NO]SO4 ,

A- EAN of Fe depends on the charge of NO ligand.
B- EAN of Fe dosen't depend upon the charge of NO.
C- Hybridisation is d2-sp3.
D- It is paramagnetic with magnetic moment= 1.73 BM.

4 Answers

29
govind ·

Ans 3 : Total number of monochlorination products = 3(without including stereochem) and 4 with stereochem

Ans 4 : B
Fe is in +1 oxidation state in this compound..and the configuration is 4s13d6..so the config will be 3d7 as one electron shifts from 4s to 3d wen the orbitals are hybridised..
3d7 so no chances of pairing..C ruled out..and it will have 3 unpaired electrons so D is also incorrect..

And EAN of this compound does not depend on the charge of NO...that's wat i think..btw NO donates three electrons in this case unlike all ligands that donate two electrons..

13
Avik ·

3) Yes, thts right govind...maine galati se galat answer dekh liya tha.

4) Ans's right; But why dosen't EAN depend upon NO's charge...?
Moreover, i didn't know NO donates 3-electrons here,,,Phir "Fe" ka Coordination no. kya hoga ??

Qn(s) 1 & 3 left too...

13
Avik ·

Okay, rest done....

To Finish off 3),
The answer is- IV>I>II>III .
Why is II>III...?

13
Avik ·

No Attempts ? :(

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